Given the following Bayes net:
with
- $p(k=t)=.2$
- $p(o=t)=.1$
- $p(s=t|k=f,o=f)=.0$
- $p(s=t|k=f,o=t)=.2$
- $p(s=t|k=t,o=f)=.5$
- $p(s=t|k=t,o=t)=.95$
how would I calculate $p(o=t|k=t)$?
Intuitively I'd say it must be $0.1$, but if I calculate this as a sum of basic probabilities: i.e. $p(s=t|o=t)=p(s=t,o=t,k=f)+p(s=t,o=t,k=t)$ the result is different.
This video also explains that it should be $0.1$, but the question explicitly said it is not $0.1$.
Your intuition is correct. Since no arrows point to either $O$ or $K$, they are independent, so $P(O=T \mid K=T)=P(O=T)=0.1$. Alternatively, you could calculate it the long way as follows: $$ \begin{align*} P(O=T \mid K=T) &= \dfrac{P(O=T,K=T)}{P(K=T)} \\ &= \dfrac{\sum_{s\in \{T,F\}}P(O=T,K=T,S=s)}{\sum_{o\in \{T,F\}}\sum_{s\in \{T,F\}}P(O=o,K=T,S=s)} \\ &= \dfrac{P(K=T)P(O=T)\sum_{s\in \{T,F\}}P(S=s \mid O=T,K=T)}{P(K=T)\sum_{o\in \{T,F\}}\sum_{s\in \{T,F\}}P(O=o)P(S=s \mid O=o,K=T)} \\ &= \dfrac{P(K=T)P(O=T)}{P(K=T)} \\ &= P(O=T) \\ &= 0.1 \end{align*} $$